Version 109 – Exam 01 – Chiu – (58295)
1
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printout
should
have
16
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
10.0 points
Two uncharged metal balls,
Z
and
X
, stand
on insulating glass rods. A third ball, carrying
a positive charge, is brought near the ball
X
as shown in the figure. A conducting wire is
then run between
Z
and
X
and then removed.
Finally the third ball is removed.
X
Z
+
conducting wire
When all this is finished
1.
ball
Z
is neutral and ball
X
is negative.
2.
ball
Z
is positive and ball
X
is neutral.
3.
balls
Z
and
X
are both negative, but ball
X
carries more charge than ball
Z
.
4.
balls
Z
and
X
are both positive.
5.
balls
Z
and
X
are both negative, but ball
Z
carries more charge than ball
X
.
6.
ball
Z
is negative and ball
X
is positive.
7.
ball
Z
is positive and ball
X
is negative.
correct
8.
ball
Z
is neutral and ball
X
is positive.
9.
ball
Z
is negative and ball
X
is neutral.
10.
balls
Z
and
X
are still uncharged.
Explanation:
When the conducting wire is run between
Z
and
X
, some positive charge flows from
X
to
Z
under the influence of the positive charge
of the third ball.
Therefore, after the wire is removed,
Z
is
charged positive and
X
is charged negative.
002
10.0 points
Two charged particles of equal magnitude
(+
Q
and +
Q
) are fixed at opposite corners
of a square that lies in a plane (see figure be
low).
A test charge
−
q
is placed at a third
corner.
+
Q
−
q
+
Q
What is the direction of the force on the
test charge due to the two other charges?
1.
2.
3.
4.
5.
correct
6.
7.
8.
Explanation:
The force between charges of the same sign
is repulsive. The force between charges with
opposite signs is attractive.
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Version 109 – Exam 01 – Chiu – (58295)
2
+
Q
−
q
+
Q
The resultant force is the sum of the two
vectors in the figure above.
Therefore the
correct choice is
003
10.0 points
Four point charges are placed at the four cor
ners of a square. Each side of the square has
a length
L
.
q
1
=
−
q
q
3
=
q
q
2
=
q
q
4
=
q
P
L
L
Find the magnitude of the electric force on
q
2
due to all three charges
q
1
,
q
3
and
q
4
. Given
L
= 1 m and
q
= 1
.
73
μ
C.
1. 0.0157246
2. 0.030333
3. 0.0446555
4. 0.0353804
5. 0.0163124
6. 0.0249351
7. 0.0143023
8. 0.0403483
9. 0.0380497
10. 0.021744
Correct answer: 0
.
0403483 N.
Explanation:
F
1
F
4y
F
4
F
3
F
4x
q
2
From the above figure, we see that
F
1
=
−
k q
2
L
2
ˆ
ı
=
−
(
8
.
98755
×
10
9
N
·
m
2
/
C
2
)
×
(1
.
73
×
10
−
6
C)
2
(1 m)
2
=
−
0
.
0268988 N
F
3
=
−
k q
2
L
2
ˆ
=
−
(
8
.
98755
×
10
9
N
·
m
2
/
C
2
)
×
(1
.
73
×
10
−
6
C)
2
(1 m)
2
=
−
0
.
0268988 N
F
4
x
=
k q
2
2
L
2
1
√
2
=
8
.
98755
×
10
9
N
·
m
2
/
C
2
2
·
(1 m)
2
×
(1
.
73
×
10
−
6
C)
2
√
2
= 0
.
00951018 N
F
4
y
=
−
k q
2
2
L
2
1
√
2
=
−
8
.
98755
×
10
9
N
·
m
2
/
C
2
2
·
(1 m)
2
×
(1
.
73
×
10
−
6
C)
2
√
2
=
−
0
.
00951018 N
bardbl
vector
F
bardbl
2
= (
F
1
+
F
4
x
)
2
+ (
F
3
+
F
4
y
)
2
= (
−
0
.
0268988 N + 0
.
00951018 N)
2
+ (
−
0
.
0268988 N +
−
0
.
00951018 N)
2
so that
bardbl
vector
F
bardbl
=
√
0
.
00162798 N
2
=
0
.
0403483 N
.
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 Spring '08
 Turner
 Physics, Charge, Electric charge

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